Explanation Approach: A matrix derived from a system of linear equations (each written in standard form with the constant term on the right) is the augmented matrix. Thus, obtain augmented matrix based on the given system of equations. Now, to apply Gauss-Jordan elimination, the augmented matrix has to be converted to the reduced row-echelon form by the elementary row transformation. Calculation: Consider the given system of equations. { x − 3 y + z = 2 3 x − y − z = − 6 − x + y − 3 z = − 2 Here, the variables are already aligned. Now, using the coefficients and constant terms as the matrix entries, the augmented matrix is, [ 1 − 3 1 ⋮ 2 3 − 1 − 1 ⋮ − 6 − 1 1 − 3 ⋮ − 2 ] Solve the system by the Gauss Jordan elimination method. Convert the augmented matrix in reduced row echelon form. Apply row operation R 2 → R 2 − 3 R 1 and R 3 → R 3 + R 1 . [ 1 − 3 1 ⋮ 2 3 − 1 − 1 ⋮ − 6 − 1 1 − 3 ⋮ − 2 ] → R 3 → R 3 + R 1 R 2 → R 2 − 3 R 1 [ 1 − 3 1 ⋮ 2 0 8 − 4 ⋮ − 12 0 − 2 − 2 ⋮ 0 ] Apply row operation R 3 → R 3 + 1 4 R 2 . [ 1 − 3 1 ⋮ 2 0 8 − 4 ⋮ − 12 0 − 2 − 2 ⋮ 0 ] → R 3 → R 3 + 1 4 R 2 [ 1 − 3 1 ⋮ 2 0 8 − 4 ⋮ − 12 0 0 − 3 ⋮ − 3 ] Apply row operation R 2 → 1 8 R 2 and R 3 → − 1 3 R 3

Bundle: Precalculus, Loose-leaf Version, 10th + WebAssign Printed Access Card for Larson's Precalculus, 10th Edition, Single-Term

10th Edition

ISBN: 9781337605090

Author: Ron Larson

Publisher: Cengage Learning

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Chapter 8.CR, Problem 28CR

To determine

To find:

The solution of the given system of linear equations based on Gauss Jordan elimination method,

{x−3y+z=23x−y−z=−6−x+y−3z=−2

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Chapter 8.CR, Problem 27CR

Chapter 8.CR, Problem 29CR

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Chapter 8 Solutions

Bundle: Precalculus, Loose-leaf Version, 10th + WebAssign Printed Access Card for Larson's Precalculus, 10th Edition, Single-Term

Ch. 8.1 - Prob. 1CP Ch. 8.1 - Prob. 2CP Ch. 8.1 - Prob. 3CP Ch. 8.1 - Prob. 4CP Ch. 8.1 - Prob. 5CP Ch. 8.1 - Prob. 6CP Ch. 8.1 - Prob. 7CP Ch. 8.1 - Prob. 8CP Ch. 8.1 - Prob. 9CP Ch. 8.1 - Prob. 1E

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The solution of the given system of linear equations based on Gauss Jordan elimination method, { x − 3 y + z = 2 3 x − y − z = − 6 − x + y − 3 z = − 2

Solution Summary: The author explains the Gauss-Jordan elimination method and how to convert the augmented matrix to the reduced row-echelon form.

Videos

To find:

The solution of the given system of linear equations based on Gauss Jordan elimination method,

{x−3y+z=23x−y−z=−6−x+y−3z=−2

Want to see the full answer?

Chapter 8 Solutions

The solution of the given system of linear equations based on Gauss Jordan elimination method, { x − 3 y + z = 2 3 x − y − z = − 6 − x + y − 3 z = − 2

Solution Summary: The author explains the Gauss-Jordan elimination method and how to convert the augmented matrix to the reduced row-echelon form.

Videos

To find: The solution of the given system of linear equations based on Gauss Jordan elimination method, {x−3y+z=23x−y−z=−6−x+y−3z=−2

Want to see the full answer?

Chapter 8 Solutions

To find:

The solution of the given system of linear equations based on Gauss Jordan elimination method,

{x−3y+z=23x−y−z=−6−x+y−3z=−2